The electric fields of two plane electromagnetic plane waves in vacuum are given by and At t = 0, a particle of charge q is at origin with a velocity v = 0.8cj (c is the speed of light in vacuum). The instantaneous force experienced by the particle is: (a) E₀ q (0.8i - j + 0.4k) (b) E₀ q (-0.8i + j + k) (c) E₀ q (0.8i + j + 0.2k) (d) E₀ q (0.4i - 3j + 0.8k)

Answer: (c). B = 1/cn × E: wave 1 along x gives B₁ = E₀/ckcos (ω t - kx); wave 2 along y gives B₂ = E₀/cicos (ω t - ky). At origin, t = 0: E = E₀(j + k), B = E₀/c(i + k). v × B = 0.8c j × E₀/c(i + k) = 0.8E₀(-k + i). F = qE₀(j + k + 0.8i - 0.8k) = E₀ q(0.8i + j + 0.2k). So the answer is (c).

The electric fields of two plane electromagnetic plane waves in… | JEE Main 2020 (9 Jan, Shift 1)

Q59JEE Main 2020 · 9 JAN · SHIFT 1Difficult

Nature of EM Waves & Wave Equation

The electric fields of two plane electromagnetic plane waves in vacuum are given by and

At t = 0, a particle of charge q is at origin with

a velocity $\vec{v} = 0.8c\hat{j}$ (c is the speed of light in vacuum). The instantaneous force experienced by the particle is:

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Answer: (c)

$\vec{B} = \dfrac{1}{c}\hat{n} \times \vec{E}$ : wave 1 along $x$ gives $\vec{B}_1 = \dfrac{E_0}{c}\hat{k}\cos(\omega t - kx)$ ; wave 2 along $y$ gives $\vec{B}_2 = \dfrac{E_0}{c}\hat{i}\cos(\omega t - ky)$ .
At origin, $t = 0$ : $\vec{E} = E_0(\hat{j} + \hat{k})$ , $\vec{B} = \dfrac{E_0}{c}(\hat{i} + \hat{k})$ .
$\vec{v} \times \vec{B} = 0.8c\,\hat{j} \times \dfrac{E_0}{c}(\hat{i} + \hat{k}) = 0.8E_0(-\hat{k} + \hat{i})$ .
$\vec{F} = qE_0(\hat{j} + \hat{k} + 0.8\hat{i} - 0.8\hat{k}) = E_0 q(0.8\hat{i} + \hat{j} + 0.2\hat{k})$ .
So the answer is (c).

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